16 Vector Calculus Copyright Cengage Learning All rights

16 Vector Calculus Copyright © Cengage Learning. All rights reserved.

16. 2 Line Integrals Copyright © Cengage Learning. All rights reserved.

Line Integrals In this section we define an integral that is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C. Such integrals are called line integrals, although “curve integrals” would be better terminology. They were invented in the early 19 th century to solve problems involving fluid flow, forces, electricity, and magnetism. 3

Line Integrals We start with a plane curve C given by the parametric equations x = x(t) y = y(t) a t b or, equivalently, by the vector equation r(t) = x(t) i + y(t) j, and we assume that C is a smooth curve. [This means that r is continuous and r (t) 0. ] 4

Line Integrals If we divide the parameter interval [a, b] into n subintervals [ti – 1, ti] of equal width and we let xi = x(ti), and yi = y(ti), then the corresponding points Pi(xi, yi) divide C into n subarcs with lengths s 1, s 2, . . . , sn. (See Figure 1. ) Figure 1 5

Line Integrals We choose any point corresponds to a point in the ith subarc. (This in [ti – 1, ti]. ) Now if f is any function of two variables whose domain includes the curve C, we evaluate f at the point multiply by the length si of the subarc, and form the sum which is similar to a Riemann sum. 6

Line Integrals Then we take the limit of these sums and make the following definition by analogy with a single integral. We have found that the length of C is 7

Line Integrals A similar type of argument can be used to show that if f is a continuous function, then the limit in Definition 2 always exists and the following formula can be used to evaluate the line integral: The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b. 8

Line Integrals If s(t) is the length of C between r(a) and r(t), then So the way to remember Formula 3 is to express everything in terms of the parameter t: Use the parametric equations to express x and y in terms of t and write ds as 9

Line Integrals In the special case where C is the line segment that joins (a, 0) to (b, 0), using x as the parameter, we can write the parametric equations of C as follows: x = x, y = 0, a x b. Formula 3 then becomes and so the line integral reduces to an ordinary single integral in this case. 10

Line Integrals Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f (x, y) 0, C f (x, y) ds represents the area of one side of the “fence” or “curtain” in Figure 2, whose base is C and whose height above the point (x, y) is f (x, y). Figure 2 11

Example 1 Evaluate C (2 + x 2 y) ds, where C is the upper half of the unit circle x 2 + y 2 = 1. Solution: In order to use Formula 3, we first need parametric equations to represent C. Recall that the unit circle can be parametrized by means of the equations x = cos t y = sin t and the upper half of the circle is described by the parameter interval 0 t . (See Figure 3. ) Figure 3 12

Example 1 – Solution cont’d Therefore Formula 3 gives 13

Line Integrals Suppose now that C is a piecewise-smooth curve; that is, C is a union of a finite number of smooth curves C 1, C 2, …. , Cn, where, as illustrated in Figure 4, the initial point of Ci + 1 is the terminal point of Ci. A piecewise-smooth curve Figure 4 14

Line Integrals Then we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C: 15

Line Integrals Any physical interpretation of a line integral C f (x, y) ds depends on the physical interpretation of the function f. Suppose that (x, y) represents the linear density at a point (x, y) of a thin wire shaped like a curve C. Then the mass of the part of the wire from Pi – 1 to Pi in Figure 1 is approximately and so the total mass of the wire is approximately Figure 1 16

Line Integrals By taking more and more points on the curve, we obtain the mass m of the wire as the limiting value of these approximations: [For example, if f (x, y) = 2 + x 2 y represents the density of a semicircular wire, then the integral in Example 1 would represent the mass of the wire. ] 17

Line Integrals The center of mass of the wire with density function is located at the point , where 18

Line Integrals Two other line integrals are obtained by replacing si by either xi = xi – 1 or yi = yi – 1 in Definition 2. They are called the line integrals of f along C with respect to x and y: 19

Line Integrals When we want to distinguish the original line integral C f (x, y) ds from those in Equations 5 and 6, we call it the line integral with respect to arc length. The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t : x = x(t), y = y(t), dx = x (t) dt, dy = y (t) dt. 20

Line Integrals It frequently happens that line integrals with respect to x and y occur together. When this happens, it’s customary to abbreviate by writing C P (x, y) dx + C Q (x, y) dy = C P (x, y) dx + Q (x, y) dy When we are setting up a line integral, sometimes the most difficult thing is to think of a parametric representation for a curve whose geometric description is given. 21

Line Integrals In particular, we often need to parametrize a line segment, so it’s useful to remember that a vector representation of the line segment that starts at r 0 and ends at r 1 is given by 22

Line Integrals In general, a given parametrization x = x(t), y = y(t), a t b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the parameter t. (See Figure 8, where the initial point A corresponds to the parameter value a and the terminal point B corresponds to t = b. ) Figure 8 23

Line Integrals If –C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A in Figure 8), then we have –C f (x, y) dx = – C f (x, y) dx –C f (x, y) dy = – C f (x, y) dy But if we integrate with respect to arc length, the value of the line integral does not change when we reverse the orientation of the curve: –C f (x, y) ds = C f (x, y) ds This is because si is always positive, whereas xi and yi change sign when we reverse the orientation of C. 24

Line Integrals in Space 25

Line Integrals in Space We now suppose that C is a smooth space curve given by the parametric equations x = x(t) y = y(t) z = z(t) a t b or by a vector equation r(t) = x(t) i + y(t) j + z(t) k. If f is a function of three variables that is continuous on some region containing C, then we define the line integral of f along C (with respect to arc length) in a manner similar to that for plane curves: 26

Line Integrals in Space We evaluate it using a formula similar to Formula 3: Observe that the integrals in both Formulas 3 and 9 can be written in the more compact vector notation For the special case f (x, y, z) = 1, we get where L is the length of the curve C. 27

Line Integrals in Space Line integrals along C with respect to x, y, and z can also be defined. For example, Therefore, as with line integrals in the plane, we evaluate integrals of the form by expressing everything (x, y, z, dx, dy, dz) in terms of the parameter t. 28

Example 5 Evaluate C y sin z ds, where C is the circular helix given by the equations x = cos t, y = sin t, z = t, 0 t 2. (See Figure 9. ) Figure 9 29

Example 5 – Solution Formula 9 gives 30

Line Integrals of Vector Fields 31

Line Integrals of Vector Fields Recall that the work done by a variable force f (x) in moving a particle from a to b along the x-axis is Then we have found that the work done by a constant force F in moving an object from a point P to another point Q in space is W = F D, where D = PQ is the displacement vector. Now suppose that F = P i + Q j + R k is a continuous force field on. (A force field on could be regarded as a special case where R = 0 and P and Q depend only on x and y. ) We wish to compute the work done by this force in moving a particle along a smooth curve C. 32

Line Integrals of Vector Fields We divide C into subarcs Pi – 1 Pi with lengths si by dividing the parameter interval [a, b] into subintervals of equal width. (See Figure 1 for the two-dimensional case or Figure 11 for the three-dimensional case. ) Figure 11 33

Line Integrals of Vector Fields Choose a point on the ith subarc corresponding to the parameter value If si is small, then as the particle moves from Pi – 1 to Pi along the curve, it proceeds approximately in the direction of the unit tangent vector at 34

Line Integrals of Vector Fields Thus the work done by the force F in moving the particle from Pi – 1 to Pi is approximately and the total work done in moving the particle along C is approximately where T(x, y, z) is the unit tangent vector at the point (x, y, z) on C. 35

Line Integrals of Vector Fields Intuitively, we see that these approximations ought to become better as n becomes larger. Therefore we define the work W done by the force field F as the limit of the Riemann sums in , namely, Equation 12 says that work is the line integral with respect to arc length of the tangential component of the force. 36

Line Integrals of Vector Fields If the curve C is given by the vector equation r(t) = x(t) i + y(t) j + z(t) k, then T(t) = r (t) / | r (t) |, so using Equation 9 we can rewrite Equation 12 in the form This integral is often abbreviated as C F dr and occurs in other areas of physics as well. 37

Line Integrals of Vector Fields Therefore we make the following definition for the line integral of any continuous vector field. When using Definition 13, remember that F(r(t)) is just an abbreviation for F(x(t), y(t), z(t)), so we evaluate F(r(t)) simply by putting x = x(t), y = y(t), and z = z(t) in the expression for F(x, y, z). Notice also that we can formally write dr = r (t) dt. 38

Example 7 Find the work done by the force field F(x, y) = x 2 i – xy j in moving a particle along the quarter-circle r(t) = cos t i + sin t j, 0 t /2. Solution: Since x = cos t and y = sin t, we have F(r(t)) = cos 2 t i – cos t sin t j and r (t) = –sin t i + cos t j 39

Example 7 – Solution cont’d Therefore the work done is 40

Line Integrals of Vector Fields Finally, we note the connection between line integrals of vector fields and line integrals of scalar fields. Suppose the vector field F on is given in component form by the equation F = P i + Q j + R k. We use Definition 13 to compute its line integral along C: 41

Line Integrals of Vector Fields But this last integral is precisely the line integral in Therefore we have . For example, the integral C y dx + z dy + x dz could be expressed as C F dr where F(x, y, z) = y i + z j + x k 42